Geometric Sequence Question (1 Viewer)

[nick90]

Gday, I have a question about Q8a(ii) of exercise 6J (pg 221) of the year 11Cambridge 3 unit maths book.

The question is to find the sum from n=3 to n=8 of log[base a] 3^n-4.

It seems simple enough, except that the second term of this series is 0, which I believe is impossible for a GP (since r=0 is invalid because then T3/T4 is indeterminable)


Anyway, the answers say the answer is 9 log[base a] 3. I can't see how they got that, and I can't see how a GP can have a 2nd term of 0.

Any help would be appreciated greatly.

 

[gman03]

log_a3^n-4 = (n-4) log_a3, right?

So sum of that from n=3 to n=8 is same as sum of (n-4) from that range, multiplied by log_a3.

Sum of n-4 from n=3 to n=8 is 9...

More like an AP question

 

[nick90]

Ah I see now! I got tricked up because it was in the GP exercise. Thanks for that.